.3 .2

1 .2 1 .3

Signed Multiply Subroutine - Single Precision

February 15, 1965

2.

ABSTRACT

This subroutine forms a 34-bit signed product from 17-bit signed multiplier and multipl icand.

3.

REQUIREMENTS

3. 1

Storage

This subroutine uses 47 (decimal) memory locations.

4.

^USAGE

4.2

Call ing Sequence

The subroutine is called by the JMS instruction. When the JMS is executed to enter the subroutine the multiplier must be in the accumulator (AC). The location following the JMS must contain a LAC with the address of the multipl icand.

The subroutine will return the instruction immediately following the latter location with the least significant part of the product in the AC. The most significant part of the product will be stored in location MP5.

6.

DESCRIPTION

Reference to the flowchart (10. 1) will illustrate the following discussion.

6. 1 • 1

On entry, the sign of the multiplier is tested, and if negative, the multiplier is made positive.

6. 1 .2 The mul tipl icand is obtained and tested for

O.

If it is found equal to 0, a

;ump to the exit is executed. Next the sign of the multiplicand is tested; and if it is found negative, the multiplicand is made positive. .

6. 1 .3 At this point, the contents of the I ink are as follows:

Sign of Multipl ier Sign of Multipl icand

o ⁰

o 1

1 0

1 1

and represent, therefore, the sign of the product.

Links

o

1

o 1

6.1.4 The multiply loop proper (tagged MP4) is entered. During this loop, the least significant half of the product shifts into the most significant end of MP5 while the multiplier shifts out the least significant end of MP5 and is lost. Note that the sign of the product is retained in MP5.

negative, complementation of the product is performed before the exit.

6.3 Scal ing

Upon entry the binary point is assumed to be located between bit positions

o

and 1 in both multiplier and multiplicand. Since there are 17 magnitude bits in each of the two factors, the product will contain 34 magnitude bits.

The product is double signed, i.e., bit positions 0 and 1 of the most sig- nificant word of the product both contain the sign. The remaining 16 bits of the most sig- nificant word of the product ~magnitude bits.

The least significant word of the product is devoted entirely to magnitude.

If the binary point of the factors are as stated above, the binary point of the product will be located between bit positions 1 and 2 in the most significant portion of the product.

On entry, multiplier and multiplicand must be 2¹s complement binary.

After return, the product is contained in two words in 2¹s complement form.

For more information on binary scal ing for fixed-point computers, see Appl ication Note 501 •

7.

METHOD

7.2

Algorithm

The conventional algorithm is used. The least significant bit of the multipl ier is tested. If it is equal to 1, the multiplicand is added to the developing product and this

quantity is shifted right. If the least significant bit of the multipLier is 0, no addition is made before the shift. The process is repeated until all the bits of the multipl ier in order from least significant to most significant have been processed.

9.

EXECUTION TIME

9. 1

Minimum

When the subroutine discovers that the multipl icand is 0, the mu Itipl ication loop is bypassed. In this case, execution time will be 14 microseconds.

9.2

Maximum

Maximum execution time occurs when the sign of the product is negative and the multipl ier consists (in binary) of all ones. The time is approximately 570 jJsec.

P D F

LIBRARY

7

10.

10. 1

CLEAR LI NK

PROGRAM Flowchart

COMPLEMENT MULTIPLIER SET LINK

COMPLEMENT MULTIPLICAND COMPLEMENT LINK

ROTATE RIGHT

g~~)p~~~7~r

ADD MULTI PLICAND C(MP5l +C(MP2l- C(ACl

STORE LEAST HALF OF PRODUCT AND MULTIPLIER C(ACl- C(MPll

COMPLEMENT BOTH HALVES OF PRODUCT

SIGN STATUS IS IN LINK. THE LAC (360000.

RAL MAKES EITHER A NOP OR CMA WHICH IS SAVED AND EXECUTED ON EXIT FROM LOOP.

SAVE MOST SIGNIFICANT HALF OF PRODUCT C (ACl -C(MP5l

The C (Y) are tested. If C (Y)

=

0, C (MP1)

=

C (MP5) = O.

If

C (Y) is not 0, then C(Y) - C(MP2), C(MP5) are cleared and multiplication is carried out as follows:

If C (MP1

)ll

contains a 1, C(MP2) are added to C (MP5). The contents of MP5 and the MPl are then snlfted right one bit. If C(MPl)17

=

0, the contents of MP5 and those of the MPl are shifted right one bit.

For this example, assume that the registers MP1, MP5 and MP2 are five bits in length instead of 17. The following sequential steps will occur in a multiply operation.

The multiplicand is 9 and the multiplier is 4.

MP5 MP1

Y

Comments

00000 01001 00100 Initial contents of the register MP1, ready to be tested.

00100 01001 C(MP2) + C(MP5) -C(MP5) since C(MPl)17 is a 1.

00010 00100 C(MP5, MP1) rotated right one place. C(MPl)17 is tested.

00001 00010 No addition, because C(MP1)lZ is O. C(MP5, MP2) rotated right one bit and AC 17 is tested.

00000 10001 No addition C(MPl)]7

=

0, C(MP5, MPl) rotated right one bit. C(MPl)17

IS

tested.

00100 10001 C (MP2) + C (MP5) C(MP5) since C(MPl )17 is a 1 •

00010 01000 C (MP5, MPl) rotated right.

00001 00100 No addition C(MP1)17

=

0, C(MP5, MPl) rotated right one bit. Rotation counter indicates that the mul tipl ication

is complete, since

it

has been reduced to o.

10.3 Program Listing

A I isting of the subroutine with MULl located at address 0200 is as follows:

/CALLING SEQUENCE:

/LAC MULTIPLIER

/JMS MULT

/LAC MULTIPLICAND

/RETURN iLOW ORDER PRODUCT IN AC

/HIGH ORDER PRODUCT IN LOCATION MP5

0200 0000 MULT, 0

0201 7100 DZM MP5 /ZERO OUT PRODUCT AREA

P D P

LIBRARY

7

0202 7510 SNA /IS MULTIPLIER ZERO?

0203 7061 JMP MPZ /IF ZERO, RETURN

0204 3250 SPA! CLL /TAKE ABSOLUTE VALUE OF MULTIPLIER 0205 3251 CMA : CML /SET LINK

=

1 IF MULTIPLIER IS NEGATIVE

0206 1600 DAC #MP1

0207 7450 XCT I MULT /PICK UP MULTIPLICAND

0207 7450 SNA /IS MULTIPLICAND ZERO

0210 5234 JMP MPZ /IF ZERO, RETURN

0211 7510 SPA /IF NON ZERO, TAKE ABSOLUTE VALUE

0212 7061 CMA ! CML /IF NEGATIVE, COMPLEMENT LINK 0213 3252 DAC #MP2 /L1NK HAS SIGN OF PRODUCT

0214 1247 LAC (360000) /COMPlEMENT ACCUMULATOR IF PRO- /DUCT IS NEGATIVE

0215 3253 RAL

0216 1250 DAC MPSIGN

0217 7010 LAM -21 /INITIALIZE COUNT TO -17

0220 3250 DAC #MP3

0221 1251 MP4, LAC MPl

0222 7430 RAR /ROTATE MULTIPLIER RIGHT ONE BIT

0223 1252 DAC MPl flOW ORDER INTO LINK

0224 7110 LAC MP5 /FETCH PRODUCT

0225 3251 SZl : Cll

0226 2253 TAD MP2 /ADD MULTIPLICAND IF LINK IS 1

0227 5216 RAR /ROTATE PRODUCT RIGHT ONE BIT

0230 1250 DAC MP5

0231 7010 ISZ MP3 /IS COUNT + 1

=

O?

0232 7430 JMP MP4 /IF NOT, GO TO MP4

0233 5240 MPSIGN, 0 /IF YES COMPLEMENT HIGH ORDER PORTION 0234 3250 DAC MP5 /OF PRODUCT, IF IT IS NEGATIVE

0235 1251 LAC MPl /RETRIEVE lOW ORDER BIT OF PRODUCT

0236 2200 RAR /FROM THE LINK

0237 5600 XCT MPSIGN /PLACE IT INTO THE LOW ORDER PORTION /OF WORD COMPLEMENT AS ABOVE

0240 7141 MPZ, ISZ MUlT

0241 3250 JMP I MULT /RETURN

STORAGE MAP: (Locations available to the user) MP5 (C(MP5)

=

high order product}